Difference between revisions of "2011 AMC 12B Problems/Problem 4"
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In multiplying two positive integers <math>a</math> and <math>b</math>, Ron reversed the digits of the two-digit number <math>a</math>. His erroneous product was <math>161.</math> What is the correct value of the product of <math>a</math> and <math>b</math>? | In multiplying two positive integers <math>a</math> and <math>b</math>, Ron reversed the digits of the two-digit number <math>a</math>. His erroneous product was <math>161.</math> What is the correct value of the product of <math>a</math> and <math>b</math>? | ||
− | <math> | + | <math>\textbf{(A)}\ 116 \qquad |
− | \textbf{(A)}\ 116 \qquad | ||
\textbf{(B)}\ 161 \qquad | \textbf{(B)}\ 161 \qquad | ||
\textbf{(C)}\ 204 \qquad | \textbf{(C)}\ 204 \qquad | ||
\textbf{(D)}\ 214 \qquad | \textbf{(D)}\ 214 \qquad | ||
\textbf{(E)}\ 224 </math> | \textbf{(E)}\ 224 </math> | ||
− | |||
== Solution == | == Solution == | ||
Line 15: | Line 13: | ||
<cmath> a*b = 32*7 = </cmath> | <cmath> a*b = 32*7 = </cmath> | ||
− | <cmath> = \boxed{224 | + | <cmath> = \boxed{224 \textbf{(E)}} </cmath> |
== See also == | == See also == | ||
− | {{AMC12 box|year=2011| | + | {{AMC12 box|year=2011|num-b=3|num-a=5|ab=B}} |
+ | {{MAA Notice}} |
Latest revision as of 14:13, 19 January 2021
Problem
In multiplying two positive integers and
, Ron reversed the digits of the two-digit number
. His erroneous product was
What is the correct value of the product of
and
?
Solution
Taking the prime factorization of reveals that it is equal to
Therefore, the only ways to represent
as a product of two positive integers is
and
Because neither
nor
is a two-digit number, we know that
and
are
and
Because
is a two-digit number, we know that a, with its two digits reversed, gives
Therefore,
and
Multiplying our two correct values of
and
yields
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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